Question

Solve the inequality -

x + 1 /( x - 1 )² < 1

Answer 1

$x +\frac{1}{(x-1)^2} < 1 \\ \\ x + \frac{1}{ {(x - 1)}^{2} } - 1 < 0 \\ \\ \frac{x {(x - 1)}^{2} + 1 - {(x - 1)}^{2} }{ {(x - 1)}^{2} } < 0 \\ \\ \frac{ {(x - 1)}^{3} + 1}{ {(x - 1)}^{2} } < 0 \\ \\ \frac{(x - 1 + 1)( {x}^{2} - 2x + 1 + 1 - x + 1) }{ {(x - 1)}^{2} } < 0 \\ \\ \frac{x( {x}^{2} - 3x + 3) }{ {(x - 1)}^{2} } < 0$
now, for (x² - 3x + 3) > 0 for all x ∈ R
because a = 1 > 0 and D = b² - 4ac = 3²-4.3<0

now,
$\frac{x(+ve)}{(x-1)^2} < 0 \\ \\ \frac{x}{ {(x - 1)}^{2} } < 0$

but we also know, (x -1)² > 0 for x ∈ R
hence,
$\frac{x}{ + ve} < 0 \\ \\ x < 0$
hence, x ∈ ( -∞ , 0)

Answer 2

★ INEQUALITIES ★

Hence , x ∈ ( - ∞ , 0 ) ∪ ( 3 , + ∞ )

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